# Wien's Law

1. Sep 8, 2013

### Hellogiraffe

1. The problem statement, all variables and given/known data
Using Wien's law, show the following:
(a) If the spectral distribution function of black body radiation, ρ(λ,T), is known at one temperature, then it can be obtained at any temperature (so that a single curve can be used to represent black body radiation at all temperatures).
(b) The total emissive power is given by R=$\sigma$$T^{4}$ (the Stefan-Boltzmann law), where $\sigma$ is a constant.
(c) The wavelength $λ_{max}$ at which ρ(λ,T) -- or R(λ,T) -- has its maximum is such that $λ_{max}$T=b (Wien's displacement law), where b is a constant.

2. Relevant equations
1) Wien's law: ρ(λ,T) = f(λT)/$λ^{5}$ with f being a function of the product of λT
2) ρ(λ,T) = (4/c)*R(λ,T)
3) R(T) = $\sigma$$T^{4}$
4) $λ_{max}$T=b

3. The attempt at a solution
So I mostly know how to derive Wien's Law from Planck's, but this question is asking to derive portions of his work without using Planck's. I got part (b) by integrating Eqn 1 from my list of relevant equations using a substitution for λT. However, I'm not even fully sure of what part (a) is even asking and where to start. For part (c), I'm assuming I'll need to take one of the forms of the functions and set its derivative (with respect to λ) equal to 0, but after substituting all different forms of the equations that I know, I cannot quite find anything that will give me what I'm looking for. I finally found another form of Wien's Law on the internet:

ρ(λ,T) = $\frac{hc^{2}}{λ^{5}}$$e^{-hc/(λkT)}$

This particular equation I was able to differentiate to give me the answer I was looking for, but there is no mention of it in the textbook I am using (Quantum Mechanics by Bransden and Joachain). If I need to, I'll just use this equation anyway, though some kind of explanation/derivation of where this came (not using Planck) from would be very helpful.

2. Sep 9, 2013

### vela

Staff Emeritus
Part (a) is saying suppose you know $\rho(\lambda,T_0)$. That is, at some temperature $T_0$, if you were given $\lambda$, you could figure out what $\rho(\lambda,T_0)$ equalled. Now suppose you have a different temperature $T_1$. You should be able to calculate what $\rho(\lambda,T_1)$ is equal to based on what you know about temperature $T_0$.

Suppose you wanted to calculate $\rho(\lambda,T_1)$. You need to figure out what $f(\lambda T_1)$ is equal to. How might you do that based on your knowledge of $\rho$ at temperature $T_0$?

That's right. Differentiate $\rho = \frac{f(\lambda T)}{\lambda^5}$ and set it equal to 0. Show that you get an equation that's a function of $\lambda T$.

3. Sep 9, 2013

### Hellogiraffe

I think the main trouble I'm having with this equation is the function as a product of two variables and how to handle it. I don't know why I'm just not seeing the link where finding $T_{0}$ will help out in finding ρ(λ,$T_{1}$). The best (and maybe only) guess I have is that I need to solve for λ since that seems to be what is fixed for all temperatures outside of the constants, but I'm not seeing a way of isolating it.

For part (c), I actually tried that earlier but I guess I'm getting really confused as to what I should do with the f(λT). I differentiated it and got to the form:

Tλf'(λT) - 5f(λT) = 0

From there, I tried doing a u-sub with u=λT, du=Tdλ. However, either my math is off or I ended up with a strange result:

$f(λT)^{T}$=A$λ^{5}$$T^{5}$

I don't have any idea what to do with that, all I really know is that I should be seeing something closer to λT=A. It's funny how I got the rest of my homework fairly easily even though they are probably harder problems, but here I am stuck on the first one :tongue2:

4. Sep 9, 2013

### vela

Staff Emeritus
Suppose you know the curve for T=100 K. In particular, say you know what $c=\rho(400\text{ nm}, 100\text{ K})$ is equal to. You have
$$\rho(400\text{ nm}, 100\text{ K}) = \frac{f(40000\text{ nm K})}{(400\text{ nm})^5} = c.$$ Now say you wanted to calculate $\rho(200\text{ nm}, 200\text{ K})$. It's equal to
$$\rho(200\text{ nm}, 200\text{ K}) = \frac{f(40000\text{ nm K})}{(200\text{ nm})^5}.$$ This quantity, I claim, is equal to $32c$. How do I know this?

In terms of $u=\lambda T$, you get $uf'(u)-5f(u)=0$. You can't explicitly solve this equation since you don't know what function $f$ is, but what variable are you solving for?

5. Sep 9, 2013

### Hellogiraffe

Ah! Finding λT was the link I was missing in (a), don't know how I didn't see that!

For (c), we are solving for u (or λT) also. I got:

f(u) = A$u^{5}$

λT = A$\sqrt[5]{f(λT)}$ with A constant

Unless f(λT) can be taken as a constant, I'm just not seeing where to take this.

6. Sep 9, 2013

### vela

Staff Emeritus
You don't want to solve it as a differential equation. In fact, since you looked up what $f(\lambda T)$ actually is, try plugging that expression into that equation. You'll see it's not satisfied.

When you try to find the maximum of $\rho(\lambda, T)$, you're trying to find a specific value $\lambda_\text{max}$ such that the condition $\rho'(\lambda_\text{max}, T)=0$ is satisfied. What you've found is that you can express this condition in terms of $u=\lambda T$. You want to explain why this implies that $\lambda_\text{max} T = b$, where $b$ is a constant.

7. Sep 9, 2013

### Hellogiraffe

I was trying way too hard to get a number and wasn't thinking it through qualitatively. Thank you so much for your help, I finally got it!